Proof of the Vector Triple Product - Practice Questions & MCQ

For three vectors $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ vector triple product is defined as $\overrightarrow{\mathbf{a}}\times (\overrightarrow{\mathbf{b}}\times \overrightarrow{\mathbf{c}})$. $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\cdot \overrightarrow{c})\cdot \overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\cdot \overrightarrow{c}$

$\overrightarrow{p}=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$ is a vector perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}\times \overrightarrow{c}$, but $\overrightarrow{b}\times \overrightarrow{c}$ is a vector perpendicular to the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$. Hence, vector $\overrightarrow{p}$ must lie in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$.
Let $\overrightarrow{p}=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=l\overrightarrow{b}+m\overrightarrow{c}[l,m$ are scalars $]$
Taking the dot product of eq (i) with $\overrightarrow{a}$, we get
$\overrightarrow{p}\cdot \overrightarrow{a}=l(\overrightarrow{a}\cdot \overrightarrow{b})+m(\overrightarrow{a}\cdot \overrightarrow{c})$

$\left[\begin{array}{l}\because \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})\text{ is }\perp \overrightarrow{a}\\ \therefore \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{a}=0\end{array}\right]$

Therefore,
$\begin{array}{ll}\Rightarrow & \overrightarrow{p}\cdot \overrightarrow{a}=0\\ \Rightarrow & l(\overrightarrow{a}\cdot \overrightarrow{b})=-m(\overrightarrow{a}\cdot \overrightarrow{c})\\ \Rightarrow & \frac{1}{\overrightarrow{a}\cdot \overrightarrow{c}}=\frac{-m}{\overrightarrow{a}\cdot \overrightarrow{b}}=\lambda \\ \Rightarrow & l=\lambda (\overrightarrow{a}\cdot \overrightarrow{c})\\ \text{ and }& m=-\lambda (\overrightarrow{a}\cdot \overrightarrow{b})\end{array}$

Substituting the value of $l$ and $m$ in Eq. (i), we get
$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\lambda [(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}]$

Here, the value of $\lambda$ can be determined by taking specific values of $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$.

The simplest way to determine $\mathit{\lambda }$ is by taking specific vectors $\overrightarrow{a}=\hat{i},\overrightarrow{b}=\hat{i},\overrightarrow{c}=\hat{j}$.
We have,
$\begin{array}{rl}& \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\lambda [(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}]\\ & \hat{i}\times (\hat{i}\times \hat{j})=\lambda [(\hat{i}\cdot \hat{j})\hat{i}-(\hat{i}\cdot \hat{i})\hat{j}]\\ & \hat{i}\times \hat{k}=\lambda [(0)\hat{i}-(1)\hat{j}]\Rightarrow -\hat{j}=-\lambda \hat{j}\\ \therefore & \lambda =1\end{array}$

Hence,
$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}$

NOTE:

1.
$\begin{array}{rl}& \overrightarrow{\mathbf{a}}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\cdot \overrightarrow{c})\cdot \overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\cdot \overrightarrow{c}\\ & (\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=(\overrightarrow{c}\cdot \overrightarrow{\mathbf{a}})\cdot \overrightarrow{b}-(\overrightarrow{c}\cdot \overrightarrow{b})\cdot \overrightarrow{a}\end{array}$
2. In general $\overrightarrow{\mathbf{a}}\times (\overrightarrow{\mathbf{b}}\times \overrightarrow{\mathbf{c}})\ne (\overrightarrow{\mathbf{a}}\times \overrightarrow{\mathbf{b}})\times \overrightarrow{\mathbf{c}}$ If $\overrightarrow{\mathbf{a}}\times (\overrightarrow{\mathbf{b}}\times \overrightarrow{\mathbf{c}})=(\overrightarrow{\mathbf{a}}\times \overrightarrow{\mathbf{b}})\times \overrightarrow{\mathbf{c}}$ then the vectors $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear.